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How do I find the common-mode rejection ratio, CMRR for this circuit? I know I have to find differential gain and common-mode gain as 'differential mode gain/commonmode gain' but with current mirrors in, I'm having trouble finding it.. Could not find though I saw other questions posted here Thank you

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    \$\begingroup\$ Hi Jessie! Can you tell us where exactly you're stuck? We can help you with problems, but we need you to narrow down for us what you need help with. Like you're currently asking, we'd need to write a complete introduction to electonic circuits, or just do your homework for you, and we'll do neither in an answer. \$\endgroup\$ – Marcus Müller Apr 9 at 12:47
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    \$\begingroup\$ Start by asking yourself what CMRR means. If this amplifier was was given to you on a PCB and you had a lab with a lot of equipment available, how would you measure the CMRR? Then do the same in the simulator. If you want to determine the CMRR analytically, then draw the small signal equivalent circuit and analyse that. Do you know how to determine the CMRR of a differential pair? Realize that if you have to ask how to determine XYZ for each existing circuit, that's not going to work. You have to understand the method so that your knowledge can be applied to ANY circuit. \$\endgroup\$ – Bimpelrekkie Apr 9 at 12:52
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CMRR = 20 * log (|Gd| / |Gcm|)

Gd = differential mode gain

Gcm = common mode gain


Let's calculate Gcm:

Add two identical voltage sources Vs1(t) to the circuit: the first on VINP and second on VINN.

We have a common mode signal applied to the circuit.

Superposition now:

Turn off (short VINN to GND) the second and calculate:

G1p(t) = VTIA'(t) / Vs1(t)   or   G1p(s) = VTIA'(s) / Vs1(s)

Now turn off (short VINP to GND) the first and calculate:

G1n(t) = VTIA''(t) / Vs1(t)   or   G1n(s) = VTIA''(s) / Vs1(s)

Now calculate Gcm:

Gcm(s) = G1n(s) + G1p(s)

or

Gcm(t) = Gn(t) + Gp(t)

Let's calculate Gd:

Add two identical voltage sources Vs2(t) to the circuit: the first on VINP and second on VINN.

This time though flip vertically Vs2(t) on VIN in order to have a differential signal applied to the circuit.

Superposition now:

Turn off (short VINN to GND) the second and calculate:

G3p(t) = VTIA'(t) / Vs2(t)   or   G2p(s) = VTIA'(s) / Vs2(s)

Turn off (short VINP to GND) the first and calculate:

Gn2(t) = VTIA''(t) / Vs2(t)   or   Gn2(s) = VTIA''(s) / Vs2(s)

Now calculate Gcm:

Gcm(s) = G2n(s) + G2p(s)

or

Gcm(t) = G2n(t) + G2p(t)

Finally calculate:

CMRR = 20 * log (|Gd| / |Gcm|)

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    \$\begingroup\$ short VINN to GND That's not how you determine CMRR. Common mode means that both the \$V_{INP}\$ and \$V_{INN}\$ are exited by the same signal meaning \$V_{INP}\$ and \$V_{INN}\$ should be shorted and a signal should be applied to them. It is good that you provide answers but giving wrong answers is only confusing to OP. \$\endgroup\$ – Bimpelrekkie Apr 9 at 17:49
  • \$\begingroup\$ You are right. I corrected my answer. \$\endgroup\$ – Enrico Migliore Apr 9 at 18:56

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